C语言10大常见基础算法-技术圈

算法是一个程序和软件的灵魂，作为一名优秀的程序员，只有对一些基础的算法有着全面的掌握，才会在设计程序和编写代码的过程中显得得心应手。本文是包括了经典的Fibonacci数列、简易计算器、回文检查、质数检查等算法。

1、计算Fibonacci数列

Fibonacci数列又称斐波那契数列，又称黄金分割数列，指的是这样一个数列：1、1、2、3、5、8、13、21。

C语言实现的代码如下：

/* Displaying Fibonacci sequence up to nth term where n is entered by user. */#include int main(){  int count, n, t1=0, t2=1, display=0;  printf("Enter number of terms: ");  scanf("%d",&n);  printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */  count=2;    /* count=2 because first two terms are already displayed. */  while (count  {      display=t1+t2;      t1=t2;      t2=display;      ++count;      printf("%d+",display);  }  return 0;}

结果输出：

Enter number of terms: 10Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+

也可以使用下面的源代码：

/* Displaying Fibonacci series up to certain number entered by user. */ #include int main(){  int t1=0, t2=1, display=0, num;  printf("Enter an integer: ");  scanf("%d",&num);  printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */  display=t1+t2;  while(display  {      printf("%d+",display);      t1=t2;      t2=display;      display=t1+t2;  }  return 0;}

结果输出：

Enter an integer: 200Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+

2、回文检查

源代码：

/* C program to check whether a number is palindrome or not */ #include int main(){  int n, reverse=0, rem,temp;  printf("Enter an integer: ");  scanf("%d", &n);  temp=n;  while(temp!=0)  {     rem=temp%10;     reverse=reverse*10+rem;     temp/=10;  }  /* Checking if number entered by user and it's reverse number is equal. */    if(reverse==n)        printf("%d is a palindrome.",n);  else      printf("%d is not a palindrome.",n);  return 0;}

结果输出：

Enter an integer: 1232112321 is a palindrome.

3、质数检查

注：1既不是质数也不是合数。

源代码：

/* C program to check whether a number is prime or not. */ #include int main(){  int n, i, flag=0;  printf("Enter a positive integer: ");  scanf("%d",&n);  for(i=2;i<=n/2;++i)  {      if(n%i==0)      {          flag=1;          break;      }  }  if (flag==0)      printf("%d is a prime number.",n);  else      printf("%d is not a prime number.",n);  return 0;}

结果输出：

Enter a positive integer: 2929 is a prime number.

4、打印金字塔和三角形

使用 * 建立三角形

** ** * ** * * ** * * * *

源代码：

#include int main(){    int i,j,rows;    printf("Enter the number of rows: ");    scanf("%d",&rows);    for(i=1;i<=rows;++i)    {        for(j=1;j<=i;++j)        {           printf("* ");        }        printf("\n");    }    return 0;}

如下图所示使用数字打印半金字塔。

11 21 2 31 2 3 41 2 3 4 5

源代码：

#include int main(){    int i,j,rows;    printf("Enter the number of rows: ");    scanf("%d",&rows);    for(i=1;i<=rows;++i)    {        for(j=1;j<=i;++j)        {           printf("%d ",j);        }        printf("\n");    }    return 0;}

用 * 打印半金字塔

* * * * ** * * ** * * * **

源代码：

#include int main(){    int i,j,rows;    printf("Enter the number of rows: ");    scanf("%d",&rows);    for(i=rows;i>=1;--i)    {        for(j=1;j<=i;++j)        {           printf("* ");        }    printf("\n");    }    return 0;}

用 * 打印金字塔

        *      * * *    * * * * *  * * * * * * ** * * * * * * * *

源代码：

#include int main(){    int i,space,rows,k=0;    printf("Enter the number of rows: ");    scanf("%d",&rows);    for(i=1;i<=rows;++i)    {        for(space=1;space<=rows-i;++space)        {           printf("  ");        }        while(k!=2*i-1)        {           printf("* ");           ++k;        }        k=0;        printf("\n");    }    return 0;}

用 * 打印倒金字塔

* * * * * * * * *  * * * * * * *    * * * * *      * * *        *

源代码：

#includeint main(){    int rows,i,j,space;    printf("Enter number of rows: ");    scanf("%d",&rows);    for(i=rows;i>=1;--i)    {        for(space=0;space           printf("  ");        for(j=i;j<=2*i-1;++j)          printf("* ");        for(j=0;j-1;++j)            printf("* ");        printf("\n");    }    return 0;}

5、简单的加减乘除计算器

源代码：

/* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */ # include int main(){    char o;    float num1,num2;    printf("Enter operator either + or - or * or divide : ");    scanf("%c",&o);    printf("Enter two operands: ");    scanf("%f%f",&num1,&num2);    switch(o) {        case '+':            printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);            break;        case '-':            printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);            break;        case '*':            printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);            break;        case '/':            printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);            break;        default:            /* If operator is other than +, -, * or /, error message is shown */            printf("Error! operator is not correct");            break;    }    return 0;}

结果输出：

Enter operator either + or - or * or divide : -Enter two operands: 3.48.43.4 - 8.4 = -5.0

6、检查一个数能不能表示成两个质数之和

源代码：

#include int prime(int n);int main(){    int n, i, flag=0;    printf("Enter a positive integer: ");    scanf("%d",&n);    for(i=2; i<=n/2; ++i)    {        if (prime(i)!=0)        {            if ( prime(n-i)!=0)            {                printf("%d = %d + %d\n", n, i, n-i);                flag=1;            }         }    }    if (flag==0)      printf("%d can't be expressed as sum of two prime numbers.",n);    return 0;}int prime(int n)      /* Function to check prime number */{    int i, flag=1;    for(i=2; i<=n/2; ++i)       if(n%i==0)          flag=0;    return flag;}

结果输出：

Enter a positive integer: 3434 = 3 + 3134 = 5 + 2934 = 11 + 2334 = 17 + 17

7、用递归的方式颠倒字符串

源代码：

/* Example to reverse a sentence entered by user without using strings. */ #include void Reverse();int main(){    printf("Enter a sentence: ");    Reverse();    return 0;}void Reverse(){    char c;    scanf("%c",&c);    if( c != '\n')    {        Reverse();        printf("%c",c);    }}

结果输出：

Enter a sentence: margorp emosewaawesome program

8、实现二进制与十进制之间的相互转换

/* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. */ #include #include int binary_decimal(int n);int decimal_binary(int n);int main(){   int n;   char c;   printf("Instructions:\n");   printf("1. Enter alphabet 'd' to convert binary to decimal.\n");   printf("2. Enter alphabet 'b' to convert decimal to binary.\n");   scanf("%c",&c);   if (c =='d' || c == 'D')   {       printf("Enter a binary number: ");       scanf("%d", &n);       printf("%d in binary = %d in decimal", n, binary_decimal(n));   }   if (c =='b' || c == 'B')   {       printf("Enter a decimal number: ");       scanf("%d", &n);       printf("%d in decimal = %d in binary", n, decimal_binary(n));   }   return 0;} int decimal_binary(int n)  /* Function to convert decimal to binary.*/{    int rem, i=1, binary=0;    while (n!=0)    {        rem=n%2;        n/=2;        binary+=rem*i;        i*=10;    }    return binary;} int binary_decimal(int n) /* Function to convert binary to decimal.*/{    int decimal=0, i=0, rem;    while (n!=0)    {        rem = n%10;        n/=10;        decimal += rem*pow(2,i);        ++i;    }    return decimal;}

结果输出：

9、使用多维数组实现两个矩阵的相加

源代码：

#include int main(){    int r,c,a[100][100],b[100][100],sum[100][100],i,j;    printf("Enter number of rows (between 1 and 100): ");    scanf("%d",&r);    printf("Enter number of columns (between 1 and 100): ");    scanf("%d",&c);    printf("\nEnter elements of 1st matrix:\n"); /* Storing elements of first matrix entered by user. */     for(i=0;i       for(j=0;j       {           printf("Enter element a%d%d: ",i+1,j+1);           scanf("%d",&a[i][j]);       } /* Storing elements of second matrix entered by user. */     printf("Enter elements of 2nd matrix:\n");    for(i=0;i       for(j=0;j       {           printf("Enter element a%d%d: ",i+1,j+1);           scanf("%d",&b[i][j]);       } /*Adding Two matrices */    for(i=0;i       for(j=0;j           sum[i][j]=a[i][j]+b[i][j]; /* Displaying the resultant sum matrix. */     printf("\nSum of two matrix is: \n\n");    for(i=0;i       for(j=0;j       {           printf("%d   ",sum[i][j]);           if(j==c-1)               printf("\n\n");       }     return 0;}

结果输出：

10、矩阵转置

源代码：

#include int main(){    int a[10][10], trans[10][10], r, c, i, j;    printf("Enter rows and column of matrix: ");    scanf("%d %d", &r, &c); /* Storing element of matrix entered by user in array a[][]. */    printf("\nEnter elements of matrix:\n");    for(i=0; i    for(j=0; j    {        printf("Enter elements a%d%d: ",i+1,j+1);        scanf("%d",&a[i][j]);    }/* Displaying the matrix a[][] */    printf("\nEntered Matrix: \n");    for(i=0; i    for(j=0; j    {        printf("%d  ",a[i][j]);        if(j==c-1)            printf("\n\n");    } /* Finding transpose of matrix a[][] and storing it in array trans[][]. */    for(i=0; i    for(j=0; j    {       trans[j][i]=a[i][j];    } /* Displaying the transpose,i.e, Displaying array trans[][]. */    printf("\nTranspose of Matrix:\n");    for(i=0; i    for(j=0; j    {        printf("%d  ",trans[i][j]);        if(j==r-1)            printf("\n\n");    }    return 0;}

结果输出：

--- EOF ---